Is ${450511}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {450511}= &&{4}\cdot100000+ \\&&{5}\cdot10000+ \\&&{0}\cdot1000+ \\&&{5}\cdot100+ \\&&{1}\cdot10+ \\&&{1}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {450511}= &&{4}(99999+1)+ \\&&{5}(9999+1)+ \\&&{0}(999+1)+ \\&&{5}(99+1)+ \\&&{1}(9+1)+ \\&&{1} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {450511}= &&\gray{4\cdot99999}+ \\&&\gray{5\cdot9999}+ \\&&\gray{0\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {4}+{5}+{0}+{5}+{1}+{1} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${450511}$ is divisible by $3$ if ${ 4}+{5}+{0}+{5}+{1}+{1}$ is divisible by $3$ Add the digits of ${450511}$ $ {4}+{5}+{0}+{5}+{1}+{1} = {16} $ If ${16}$ is divisible by $3$ , then ${450511}$ must also be divisible by $3$ ${16}$ is not divisible by $3$, therefore ${450511}$ must not be divisible by $3$.